Polynomials

Hong Kong

Stage 4 - Stage 5

Lesson

A zero of a polynomial is an $x$`x`-value that when substituted into the polynomial gives the value of $0$0.

A simple example, the polynomial $g(x)=x-4$`g`(`x`)=`x`−4, has a zero of $x=4$`x`=4.

When we substitute $x=4$`x`=4 into the polynomial the answer is $0$0. $g(4)=4-4=0$`g`(4)=4−4=0.

Finding zeros will require us to use our skills in algebraic manipulation and factorisation.

A polynomial $g(x)=x^2+kx-28$`g`(`x`)=`x`2+`k``x`−28 has a zero at $x=4$`x`=4. Find the other zero.

**Think**: If $x=4$`x`=4 is a zero of the polynomial $g(x)$`g`(`x`). Then when we substitute $x=4$`x`=4 into the polynomial the answer will be zero. Let's see what happens when we do that here.

$g(x)$g(x) |
$=$= | $x^2+kx-28$x2+kx−28 |

$g(4)$g(4) |
$=$= | $4^2+k\times4-28$42+k×4−28 |

$g(4)$g(4) |
$=$= | $16+4k-28$16+4k−28 |

$g(4)$g(4) |
$=$= | $-12+4k$−12+4k |

At this stage we can solve for $k$`k`, because we know that $x=4$`x`=4 is a zero and thus $g(4)=0$`g`(4)=0.

Therefore $-12+4k$−12+4`k` equals $0$0. Hence $-12+4k=0$−12+4`k`=0, so $k=3$`k`=3.

We can now express the polynomial as $g(x)=x^2+3x-28$`g`(`x`)=`x`2+3`x`−28, but we are yet to solve for the second zero.

How can we do that? Well we know that to solve a quadratic we can factorise fully. Once factorised each of the linear factors present us with a solution. In this case we already know that $x=4$`x`=4 is a solution, this means that $x-4$`x`−4 is one of the linear factors.

Factorising $g(x)$`g`(`x`), we get $g(x)=(x-4)(x+7)$`g`(`x`)=(`x`−4)(`x`+7), which means that the other zero is at $x+7=0$`x`+7=0, which is $x=-7$`x`=−7.

What we have done here was find the value $k$`k` to get the complete polynomial, and then we were able to factorise and solve to find the other zero.

Knowing at least one of the zeros is always a good start, as from this we have a factor we can use to begin our factorisation. From there we need to finish factorising our polynomial and the solve to find the other zeros. Sometimes you can factorise by inspection (especially if the function is linear or quadratic), and sometimes you will need to use polynomial division. To remind yourself of some of these skills, these sections might be helpful.

- Highest common factor
- Complete the square
- Grouping in pairs
- Special factorisations like sum and difference of cubes, difference of two squares, perfect squares.
- Polynomial division and synthetic division.

Remember, that values of the zeros are the same as the values of the roots of the equation. We are finding where the function crosses the $x$`x`-axis. The highest power of the function tells you how many (at most) roots (or zeros) the polynomial might have.

The polynomial $P\left(x\right)=x^2+kx+8$`P`(`x`)=`x`2+`k``x`+8 has a zero at $x=4$`x`=4.

Solve for the other zero of $P\left(x\right)$

`P`(`x`).Solve for the value of $k$

`k`.

The quadratic polynomial $P\left(x\right)=x^2+k-4x$`P`(`x`)=`x`2+`k`−4`x` has equal zeros.

Find the value of the equal zeros.

Solve for the value of $k$

`k`.

The polynomial $P\left(x\right)=x^3-5x^2+5x+3$`P`(`x`)=`x`3−5`x`2+5`x`+3 has a zero at $x=3$`x`=3. Solve for the other zeros of $P\left(x\right)$`P`(`x`).